Question

1.According to the ideal gas law, a 1.052 mol sample of methane gas in a 1.031 L container at 271.3 K should exert a pressure of 22.72 atm. By what percent does the pressure calculated using the van der Waals' equation differ from the ideal pressure? For CH4 gas,
a = 2.253 L2atm/mol2 and
b = 4.278×10-2 L/mol.

Answer 1

Answer:

They differ in a 5.86%

Explanation:

Hi, the Van der Waals equation for gases is:

$(P+ \frac{a}{v^2})(v-b)=R*T$

For methane gas:

$a= 2.253 \frac{L^2atm}{mol^2}$

$b= 4.278*10^{-2} \frac{L}{mol}$

The conditions give{n are T=271.3 K, n=1.052mol and V=1.031L

The molar volume:

$v= \frac{V}{n}$

$v= \frac{1.031L}{1.052mol}$

$v= 0.98 \frac{L}{mol}$

Replacing in Van der Waals :

$(P+ \frac{2.253 \frac{L^2atm}{mol^2}}{ 0.98 \frac{L}{mol}^2})( 0.98 \frac{L}{mol}-4.278×10^-2 \frac{L}{mol})=0.082 \frac{L*atm}{mol*K}*271.3K$

$P=21.39 atm$

In percentaje:

$\frac{P_{VdW}}{P{ideal}}*100=\frac{21.39 atm}{22.72atm}*100$

$\frac{P_{VdW}}{P{ideal}}*100=94.14$

They differ in a 5.86%